Hi everyone.
Show that $\displaystyle \forall{x\in{\mathbb{Z^+}\cup{0}}}$ then $\displaystyle [\frac{n}{2}]+[-\frac{n}{2}]=n$
Thanks so much.
You mean $\displaystyle \forall n\in\mathbb Z^+\cup\{0\},$ $\displaystyle \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=n?$
If $\displaystyle n$ is even, say $\displaystyle n=2k,$ $\displaystyle \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=k+\left|-k\right|=2k=n.$
If $\displaystyle n$ is odd, say $\displaystyle n=2k+1,$ $\displaystyle \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=\lfloor k+\tfrac12\rfloor+\left|\lfloor-k-\tfrac12\rfloor\right|=k+|-k-1|=2k+1=n.$